Beregn summen af alle tal i en streng
Prøv det på GfG Practice 
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Produktion
Produktion
Produktion
#practiceLinkDiv { display: ingen !important; } Givet en streng S indeholdende alfanumeriske tegn Opgaven er at beregne summen af alle tal, der findes i strengen.
Eksempler:
Anbefalet praksis Summen af tal i streng Prøv det!Input: 1abc23
Produktion: 24
Forklaring: 1 + 23 = 24
Input: nørder 4 nørder
Produktion: 4Input: 1abc2x30yz67
Produktion: 100
Nærme sig:
Scan hvert tegn i inputstrengen, og hvis et tal er dannet af fortløbende tegn i strengen, skal du øge resultat med det beløb. Den eneste vanskelige del af dette spørgsmål er, at flere på hinanden følgende cifre betragtes som ét tal.
Følg nedenstående trin for at implementere ideen:
- Opret en tom streng temp og et heltal sum .
- Gentag over alle tegn i strengen.
- Hvis tegnet er et numerisk ciffer, føjes det til temp .
- Ellers konverter midlertidig streng til nummer og føj den til sum tom temp .
- Retursum + tal opnået fra temp.
Nedenfor er implementeringen af ovenstående tilgang:
C++ // C++ program to calculate sum of all numbers present // in a string containing alphanumeric characters #include using namespace std ; // Function to calculate sum of all numbers present // in a string containing alphanumeric characters int findSum ( string str ) { // A temporary string string temp = '' ; // holds sum of all numbers present in the string int sum = 0 ; // read each character in input string for ( char ch : str ) { // if current character is a digit if ( isdigit ( ch )) temp += ch ; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += atoi ( temp . c_str ()); // reset temporary string to empty temp = '' ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + atoi ( temp . c_str ()); } // Driver code int main () { // input alphanumeric string string str = '12abc20yz68' ; // Function call cout < < findSum ( str ); return 0 ; }
Java // Java program to calculate sum of all numbers present // in a string containing alphanumeric characters import java.io.* ; class GFG { // Function to calculate sum of all numbers present // in a string containing alphanumeric characters static int findSum ( String str ) { // A temporary string String temp = '0' ; // holds sum of all numbers present in the string int sum = 0 ; // read each character in input string for ( int i = 0 ; i < str . length (); i ++ ) { char ch = str . charAt ( i ); // if current character is a digit if ( Character . isDigit ( ch )) temp += ch ; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += Integer . parseInt ( temp ); // reset temporary string to empty temp = '0' ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + Integer . parseInt ( temp ); } // Driver code public static void main ( String [] args ) { // input alphanumeric string String str = '12abc20yz68' ; // Function call System . out . println ( findSum ( str )); } } // This code is contributed by AnkitRai01
Python3 # Python3 program to calculate sum of # all numbers present in a string # containing alphanumeric characters # Function to calculate sum of all # numbers present in a string # containing alphanumeric characters def findSum ( str1 ): # A temporary string temp = '0' # holds sum of all numbers # present in the string Sum = 0 # read each character in input string for ch in str1 : # if current character is a digit if ( ch . isdigit ()): temp += ch # if current character is an alphabet else : # increment Sum by number found # earlier(if any) Sum += int ( temp ) # reset temporary string to empty temp = '0' # atoi(temp.c_str1()) takes care # of trailing numbers return Sum + int ( temp ) # Driver code # input alphanumeric string str1 = '12abc20yz68' # Function call print ( findSum ( str1 )) # This code is contributed # by mohit kumar
C# // C# program to calculate sum of // all numbers present in a string // containing alphanumeric characters using System ; class GFG { // Function to calculate sum of // all numbers present in a string // containing alphanumeric characters static int findSum ( String str ) { // A temporary string String temp = '0' ; // holds sum of all numbers // present in the string int sum = 0 ; // read each character in input string for ( int i = 0 ; i < str . Length ; i ++ ) { char ch = str [ i ]; // if current character is a digit if ( char . IsDigit ( ch )) temp += ch ; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += int . Parse ( temp ); // reset temporary string to empty temp = '0' ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + int . Parse ( temp ); } // Driver code public static void Main ( String [] args ) { // input alphanumeric string String str = '12abc20yz68' ; // Function call Console . WriteLine ( findSum ( str )); } } // This code is contributed by PrinciRaj1992
JavaScript < script > // Javascript program to calculate // sum of all numbers present // in a string containing // alphanumeric characters // Function to calculate sum // of all numbers present // in a string containing // alphanumeric characters function findSum ( str ) { // A temporary string let temp = '0' ; // holds sum of all numbers // present in the string let sum = 0 ; // read each character in input string for ( let i = 0 ; i < str . length ; i ++ ) { let ch = str [ i ]; // if current character is a digit if ( ! isNaN ( String ( ch ) * 1 )) temp += ch ; // if current character is an alphabet else { // increment sum by number found earlier // (if any) sum += parseInt ( temp ); // reset temporary string to empty temp = '0' ; } } // atoi(temp.c_str()) takes care of trailing // numbers return sum + parseInt ( temp ); } // Driver code // input alphanumeric string let str = '12abc20yz68' ; // Function call document . write ( findSum ( str )); // This code is contributed by unknown2108 < /script>
Produktion
100
Tidskompleksitet: O(N) hvor n er længden af strengen.
Hjælpeplads: O(N) hvor n er længden af strengen.
Beregn summen af alle tal til stede i en streng vha rekursion
Ideen er rekursivt at krydse over strengen og finde ud af tal læg derefter disse tal til resultat til sidst returnere resultat .
Følg nedenstående trin for at implementere ideen:
- Opret en tom streng temp og et heltal sum .
- Gennemgå rekursivt tegnene for hvert indeks jeg fra til længde - 1 .
- Hvis i = N-1 kontroller derefter, om det aktuelle tegn er et cifferretur str[i] - '0' .
- Ellers vende tilbage .
- Hvis str[i] er et ciffer.
- Kør en for-løkke med tæller j fra jeg til N - 1 .
- Hvis tegnet er et numerisk ciffer, føjes det til temp .
- Ellers pause.
- Retur sum af numerisk værdi af temp + tilbagevendende for indeks j .
- Kør en for-løkke med tæller j fra jeg til N - 1 .
Nedenfor er implementeringen af ovenstående tilgang:
C++ // C++ program to calculate sum of all numbers // present in a string containing alphanumeric // characters #include using namespace std ; int solve ( string & str int i int n ) { // if string is empty if ( i >= n ) return 0 ; // if on the last index if ( i == n - 1 ) { // if last digit is numeric if ( isdigit ( str [ i ])) { return str [ i ] - '0' ; } else { return 0 ; } } // if current char is digit // then sum the consecutive digits if ( isdigit ( str [ i ])) { // declared an empty string string temp = '' ; int j ; // start from that index // sum all the consecutive digits for ( j = i ; j < n ; j ++ ) { // if current char is digit // add it to the temp string if ( isdigit ( str [ j ])) temp += str [ j ]; // if it is not a digit // break instantly else break ; } // add the number associated to temp // with the answer recursion will bring return stoi ( temp ) + solve ( str j n ); } // else call from the next index else { solve ( str i + 1 n ); } } int findSum ( string str ) { // recursiven function return solve ( str 0 str . size ()); } // Driver code int main () { // input alphanumeric string string str = '12abc20yz68' ; // Function call cout < < findSum ( str ); return 0 ; }
Java import java.util.Scanner ; class Main { static int solve ( String str int i int n ) { // if string is empty if ( i >= n ) return 0 ; // if on the last index if ( i == n - 1 ) { // if last digit is numeric if ( Character . isDigit ( str . charAt ( i ))) { return str . charAt ( i ) - '0' ; } else { return 0 ; } } // if current char is digit // then sum the consecutive digits if ( Character . isDigit ( str . charAt ( i ))) { // declared an empty string String temp = '' ; int j ; // start from that index // sum all the consecutive digits for ( j = i ; j < n ; j ++ ) { // if current char is digit // add it to the temp string if ( Character . isDigit ( str . charAt ( j ))) temp += str . charAt ( j ); // if it is not a digit // break instantly else break ; } // add the number associated to temp // with the answer recursion will bring return Integer . parseInt ( temp ) + solve ( str j n ); } // else call from the next index else { return solve ( str i + 1 n ); } } static int findSum ( String str ) { // recursiven function return solve ( str 0 str . length ()); } // Driver code public static void main ( String [] args ) { // input alphanumeric string String str = '12abc20yz68' ; // Function call System . out . println ( findSum ( str )); } } // This code contributed by Ajax
Python3 def findSum ( str ): # variable to store sum result = 0 temp = '' for i in range ( len ( str )): if str [ i ] . isnumeric (): temp += str [ i ] if i == len ( str ) - 1 : result += int ( temp ) else : if temp != '' : result += int ( temp ) temp = '' return result # driver code if __name__ == '__main__' : # input alphanumeric string str = '12abc20yz68' print ( findSum ( str )) #This code contributed by Shivam Tiwari
C# // C# program to calculate sum of all numbers // present in a string containing alphanumeric // characters using System ; using System.Linq ; using System.Collections.Generic ; class GFG { static bool isdigit ( char c ) { if ( c >= '0' && c <= '9' ) return true ; return false ; } static int solve ( string str int i int n ) { // if string is empty if ( i >= n ) return 0 ; // if on the last index if ( i == n - 1 ) { // if last digit is numeric if ( isdigit ( str [ i ])) { return str [ i ]; } else { return 0 ; } } // if current char is digit // then sum the consecutive digits if ( isdigit ( str [ i ])) { // declared an empty string string temp = '' ; int j ; // start from that index // sum all the consecutive digits for ( j = i ; j < n ; j ++ ) { // if current char is digit // add it to the temp string if ( isdigit ( str [ j ])) temp += str [ j ]; // if it is not a digit // break instantly else break ; } // add the number associated to temp // with the answer recursion will bring return Int32 . Parse ( temp ) + solve ( str j n ); } // else call from the next index else { return solve ( str i + 1 n ); } } static int findSum ( string str ) { // recursiven function return solve ( str 0 str . Length ); } // Driver code static public void Main () { // input alphanumeric string string str = '12abc20yz68' ; // Function call Console . Write ( findSum ( str )); } }
JavaScript function findSum ( str ) { // variable to store sum let result = 0 ; let temp = '' ; for ( let i = 0 ; i < str . length ; i ++ ) { if ( ! isNaN ( str [ i ])) { temp += str [ i ]; if ( i === str . length - 1 ) { result += parseInt ( temp ); } } else { if ( temp !== '' ) { result += parseInt ( temp ); temp = '' ; } } } return result ; } // driver code console . log ( findSum ( '12abc20yz68' )); // This code is contributed by Shivam Tiwari
Produktion
100
Tidskompleksitet: PÅ) hvor N er størrelsen af den givne streng.
Hjælpeplads: PÅ) i værste fald kan det koste O(N) rekursive opkald
Beregn summen af alle tal til stede i en streng ved hjælp af Regex i Python:
Ideen er at bruge indbygget funktion Python RegEx .
Nedenfor er implementeringen af ovenstående tilgang:
C++14 #include #include // Function to calculate sum of all // numbers present in a string // containing alphanumeric characters int findSum ( std :: string str ) { // Regular Expression that matches // digits in between a string std :: regex pattern ( ' \ d+' ); std :: smatch match ; int sum = 0 ; while ( std :: regex_search ( str match pattern )) { sum += stoi ( match [ 0 ]. str ()); str = match . suffix (). str (); } return sum ; } // Driver code int main () { // input alphanumeric string std :: string str = '12abc20yz68' ; // Function call std :: cout < < findSum ( str ) < < std :: endl ; return 0 ; } // This code is contributed by Shivam Tiwari
Python3 # Python3 program to calculate sum of # all numbers present in a string # containing alphanumeric characters # Function to calculate sum of all # numbers present in a string # containing alphanumeric characters import re def find_sum ( str1 ): # Regular Expression that matches # digits in between a string return sum ( map ( int re . findall ( 'd+' str1 ))) # Driver code # input alphanumeric string str1 = '12abc20yz68' # Function call print ( find_sum ( str1 )) # This code is contributed # by Venkata Ramana B
JavaScript // JavaScript program to calculate sum of // all numbers present in a string // containing alphanumeric characters // Function to calculate sum of all // numbers present in a string // containing alphanumeric characters function find_sum ( str1 ) { // Regular Expression that matches // digits in between a string return str1 . match ( /d+/g ). reduce (( acc val ) => acc + parseInt ( val ) 0 ); } // Driver code // input alphanumeric string const str1 = '12abc20yz68' ; // Function call console . log ( find_sum ( str1 ));
Java import java.util.regex.* ; public class Main { // Function to calculate sum of all // numbers present in a string // containing alphanumeric characters public static int findSum ( String str ) { // Regular Expression that matches // digits in between a string Pattern pattern = Pattern . compile ( '\d+' ); Matcher matcher = pattern . matcher ( str ); int sum = 0 ; while ( matcher . find ()) { sum += Integer . parseInt ( matcher . group ()); str = matcher . replaceFirst ( '' ); matcher = pattern . matcher ( str ); } return sum ; } // Driver code public static void main ( String [] args ) { // input alphanumeric string String str = '12abc20yz68' ; // Function call System . out . println ( findSum ( str )); } }
C# using System ; using System.Text.RegularExpressions ; public class GFG { // Function to calculate sum of all // numbers present in a string // containing alphanumeric characters public static int FindSum ( string str ) { // Regular Expression that matches // digits in between a string Regex pattern = new Regex ( @'d+' ); Match matcher = pattern . Match ( str ); int sum = 0 ; while ( matcher . Success ) { sum += Int32 . Parse ( matcher . Value ); str = pattern . Replace ( str '' 1 matcher . Index ); matcher = pattern . Match ( str ); } return sum ; } // Main method static public void Main () { // input alphanumeric string string str = '12abc20yz68' ; // Function call Console . WriteLine ( FindSum ( str )); } }
Produktion
100
Tidskompleksitet: O(n) hvor n er længden af strengen.
Hjælpeplads: O(n) hvor n er længden af strengen.
